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In △ABC, CM is the median, CD is the angle bisector of ∠ACB, CH is the altitude. CM, CD, and CH divide ∠ACB by four equal angles. Find angles of △ABC.

I started with finding angle relationships but was not able to use them to proceed. Maybe this is not the right approach. Can you advise? Thanks.

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2 Answers 2

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Assume that $AC<BC$. Let $O$ be the centre of the circumcircle of $ABC$. Prove that (in any triangle) angles $\sphericalangle{BCO}=\sphericalangle{ACH}$. Since the angle bisector already divides the angle in halves, we have to have that angle $\sphericalangle{ACH}$ is a quater of angle $C$ as well as $\sphericalangle{BCM}$. But since $\sphericalangle{BCO}=\sphericalangle{ACH}$ we must have $\sphericalangle{BCO}=\sphericalangle{BCM}$ i.e. points $C$, $M$, $O$ are collinear. Prove that it implies that $\sphericalangle{C}=90^{\circ}$ and usethe fact that then $\sphericalangle{ACH}=\frac{90^{\circ}}{4}$ to compute the rest.

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Let $\measuredangle ACH=\alpha$.

Thus, $\measuredangle A=90^{\circ}-\alpha$ and $\measuredangle B=90^{\circ}-3\alpha.$

Thus, by the law of sines we obtain: $$\frac{\sin3\alpha}{\cos\alpha}=\frac{AM}{MC}=\frac{BM}{MC}=\frac{\sin\alpha}{\cos3\alpha},$$ which gives $$\sin6\alpha=\sin2\alpha$$ or $$6\alpha+2\alpha=180^{\circ}.$$ Can you end it now?

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